Kummer's Test

Weighting the ratio by a positive auxiliary sequence (pn)(p_n), the sign of Kn=pnanan+1−pn+1K_n = p_n\frac{a_n}{a_{n+1}} - p_{n+1} decides convergence. pn=1p_n = 1 recovers the ratio test, pn=np_n = n recovers Raabe's test.

Kn:=pnanan+1−pn+1≥c>0 ⇒ ∑an convergesK_n := p_n\frac{a_n}{a_{n+1}} - p_{n+1} \ge c > 0 \ \Rightarrow\ \sum a_n\ \text{converges}
Kn≤0, ∑1pn=∞ ⇒ ∑an divergesK_n \le 0,\ \sum\frac{1}{p_n} = \infty \ \Rightarrow\ \sum a_n\ \text{diverges}

Part of these learning paths

Examples & Exercises

Prove Kummer's test: let an>0a_n > 0, pn>0p_n > 0, and Kn:=pnanan+1−pn+1K_n := p_n \frac{a_n}{a_{n+1}} - p_{n+1}. (i) If Kn≥c>0K_n \ge c > 0 eventually, then ∑an\sum a_n converges. (ii) If Kn≤0K_n \le 0 eventually and ∑1pn\sum \frac{1}{p_n} diverges, then ∑an\sum a_n diverges.

Kn≥c>0⟹c an+1≤pnan−pn+1an+1K_n \ge c > 0 \quad \Longrightarrow \quad c\, a_{n+1} \le p_n a_n - p_{n+1} a_{n+1}
Kn≤0⟹pnan≤pn+1an+1⟹an≥dpnK_n \le 0 \quad \Longrightarrow \quad p_n a_n \le p_{n+1} a_{n+1} \quad \Longrightarrow \quad a_n \ge \frac{d}{p_n}
pn=np_n = n, a=2a = 2, n=100n = 100: K1=0K_1 = 0, K2=13K_2 = \frac13, K100=99101≈0.9802K_{100} = \frac{99}{101} \approx 0.9802 — Kn≥12>0K_n \ge \frac12 > 0 for n≥3n \ge 3, so the series converges.
Show proof
  1. Convergent half: suppose Kn≥c>0K_n \ge c > 0 for n≥Nn \ge N. Multiplying Kn=pnanan+1−pn+1≥cK_n = p_n \frac{a_n}{a_{n+1}} - p_{n+1} \ge c by the positive an+1a_{n+1} gives pnan−pn+1an+1≥c an+1p_n a_n - p_{n+1} a_{n+1} \ge c\, a_{n+1}.
  2. Summing from n=Nn = N to MM, the right side telescopes: c∑n=NMan+1≤pNaN−pM+1aM+1≤pNaNc \sum_{n=N}^{M} a_{n+1} \le p_N a_N - p_{M+1} a_{M+1} \le p_N a_N (the last inequality uses pM+1aM+1>0p_{M+1} a_{M+1} > 0). So the partial sums of the positive-term series ∑an\sum a_n are bounded above: ∑an\sum a_n converges.
  3. Divergent half: suppose Kn≤0K_n \le 0 for n≥Nn \ge N. Multiplying pnanan+1≤pn+1p_n \frac{a_n}{a_{n+1}} \le p_{n+1} by the positive an+1a_{n+1} gives pnan≤pn+1an+1p_n a_n \le p_{n+1} a_{n+1} — so pnanp_n a_n is increasing, and pnan≥pNaN=:d>0p_n a_n \ge p_N a_N =: d > 0 for n≥Nn \ge N.
  4. Hence an≥dpna_n \ge \frac{d}{p_n} for n≥Nn \ge N, and since ∑1pn\sum \frac{1}{p_n} diverges, comparison gives that ∑an\sum a_n diverges. The intuition: pnp_n is a probe, and the ratio test (pn=1p_n = 1) and Raabe's test (pn=np_n = n) are its special cases.
  5. Numerical check (pn=np_n = n, a=2a = 2): Kn=n−1n+1K_n = \frac{n-1}{n+1}, so K1=0K_1 = 0, K2=13K_2 = \frac13, K100=99101≈0.9802K_{100} = \frac{99}{101} \approx 0.9802 — matching the caption, and Kn≥12K_n \ge \frac12 for n≥3n \ge 3 puts the convergent half to work with c=12c = \frac12.

Show that Raabe's test is Kummer's test with pn=np_n = n, and prove that for L:=lim⁡n→∞n(anan+1−1)L := \lim_{n\to\infty} n\left(\frac{a_n}{a_{n+1}} - 1\right): convergence if L>1L > 1, divergence if L<1L < 1.

pn=n:Kn=n(anan+1−1)−1  →n→∞  L−1p_n = n: \qquad K_n = n\left(\frac{a_n}{a_{n+1}} - 1\right) - 1 \;\xrightarrow[n\to\infty]{}\; L - 1
Gauss example a=2a = 2: n(anan+1−1)=2nn+1→L=2n\left(\frac{a_n}{a_{n+1}} - 1\right) = \frac{2n}{n+1} \to L = 2, K100=99101≈0.9802→L−1=1K_{100} = \frac{99}{101} \approx 0.9802 \to L - 1 = 1 — convergent since L>1L > 1.
Show proof
  1. Plugging in pn=np_n = n gives Kn=nanan+1−(n+1)=n(anan+1−1)−1K_n = n \frac{a_n}{a_{n+1}} - (n+1) = n\left(\frac{a_n}{a_{n+1}} - 1\right) - 1. Taking limits, Kn→L−1K_n \to L - 1 — the Raabe quotient shifted by 1.
  2. If L>1L > 1, then Kn→L−1>0K_n \to L - 1 > 0, so eventually Kn≥L−12>0K_n \ge \frac{L-1}{2} > 0. Kummer's convergent half (with c=L−12c = \frac{L-1}{2}) gives convergence of ∑an\sum a_n.
  3. If L<1L < 1, then Kn→L−1<0K_n \to L - 1 < 0, so eventually Kn≤L−12<0K_n \le \frac{L-1}{2} < 0. And ∑1pn=∑1n\sum \frac{1}{p_n} = \sum \frac{1}{n} diverges (harmonic series), so Kummer's divergent half gives divergence of ∑an\sum a_n.
  4. Numerical check (Gauss example a=2a = 2): anan+1=1+2n+1\frac{a_n}{a_{n+1}} = 1 + \frac{2}{n+1}, so n(anan+1−1)=2nn+1→2=Ln\left(\frac{a_n}{a_{n+1}} - 1\right) = \frac{2n}{n+1} \to 2 = L and Kn=2nn+1−1=n−1n+1→1=L−1K_n = \frac{2n}{n+1} - 1 = \frac{n-1}{n+1} \to 1 = L - 1 — K100=99101≈0.9802K_{100} = \frac{99}{101} \approx 0.9802 matches the caption.

As a corollary of Gauss's test: for an=n!(a+1)(a+2)⋯(a+n)a_n = \frac{n!}{(a+1)(a+2)\cdots(a+n)}, show anan+1=1+an+O(1n2)\frac{a_n}{a_{n+1}} = 1 + \frac{a}{n} + O\left(\frac{1}{n^2}\right) (i.e. Gauss h=ah = a), then prove via Kummer's test that ∑an\sum a_n converges for a=2a = 2 and diverges for a=12a = \frac12.

anan+1=n+a+1n+1=1+an+1=1+an+O(1n2),h=a\frac{a_n}{a_{n+1}} = \frac{n+a+1}{n+1} = 1 + \frac{a}{n+1} = 1 + \frac{a}{n} + O\left(\frac{1}{n^2}\right), \qquad h = a
a=2:Kn=n−1n+1→1>0    (converges),a=12:Kn=−0.5n−1n+1→−12<0    (diverges)a = 2: \quad K_n = \frac{n-1}{n+1} \to 1 > 0 \;\; (\text{converges}), \qquad a = \tfrac12: \quad K_n = \frac{-0.5n-1}{n+1} \to -\tfrac12 < 0 \;\; (\text{diverges})
pn=np_n = n, a=12a = \frac12, n=100n = 100: K1=−0.75K_1 = -0.75, K100=−51101≈−0.505<0K_{100} = \frac{-51}{101} \approx -0.505 < 0 — the series diverges since ∑1n\sum \frac1n diverges. (At the same nn, a=2a = 2 gives K100=99101≈0.9802>0K_{100} = \frac{99}{101} \approx 0.9802 > 0: convergent.)
Show proof
  1. Computing the ratio directly: anan+1=n!(a+1)⋯(a+n)⋅(a+1)⋯(a+n+1)(n+1)!=n+a+1n+1=1+an+1\frac{a_n}{a_{n+1}} = \frac{n!}{(a+1)\cdots(a+n)} \cdot \frac{(a+1)\cdots(a+n+1)}{(n+1)!} = \frac{n+a+1}{n+1} = 1 + \frac{a}{n+1}. Since an+1=an⋅nn+1=an−an(n+1)\frac{a}{n+1} = \frac{a}{n} \cdot \frac{n}{n+1} = \frac{a}{n} - \frac{a}{n(n+1)}, this is the Gauss form 1+hn+O(1n2)1 + \frac{h}{n} + O\left(\frac{1}{n^2}\right) with h=ah = a. And an+1→0\frac{a}{n+1} \to 0, so the ratio tends to 11: the ratio test is powerless.
  2. For a=2a = 2 we verify the closed form: an=n!3⋅4⋯(n+2)=n!⋅2(n+2)!=2(n+1)(n+2)a_n = \frac{n!}{3 \cdot 4 \cdots (n+2)} = \frac{n! \cdot 2}{(n+2)!} = \frac{2}{(n+1)(n+2)} (multiplying top and bottom by 1⋅2=21 \cdot 2 = 2 completes (n+2)!(n+2)!). Indeed a1=13=22⋅3a_1 = \frac{1}{3} = \frac{2}{2\cdot 3}. So an=2(1n+1−1n+2)a_n = 2\left(\frac{1}{n+1} - \frac{1}{n+2}\right) telescopes: ∑n=1Man=2(12−1M+2)→1\sum_{n=1}^{M} a_n = 2\left(\frac12 - \frac{1}{M+2}\right) \to 1 — the series sums to 11.
  3. Back to the Kummer quotient (pn=np_n = n): Kn=nn+a+1n+1−(n+1)=(a−1)n−1n+1→a−1K_n = n\frac{n+a+1}{n+1} - (n+1) = \frac{(a-1)n - 1}{n+1} \to a - 1. For a=2a = 2, Kn=n−1n+1→1>0K_n = \frac{n-1}{n+1} \to 1 > 0 (with Kn≥12K_n \ge \frac12 for n≥3n \ge 3), so Kummer's convergent half gives convergence — agreeing with the telescoping closed form (K100=99101≈0.9802K_{100} = \frac{99}{101} \approx 0.9802).
  4. For a=12a = \frac12, Kn=−0.5n−1n+1→−12<0K_n = \frac{-0.5n-1}{n+1} \to -\frac12 < 0, so eventually Kn≤0K_n \le 0. Since ∑1pn=∑1n\sum \frac{1}{p_n} = \sum \frac{1}{n} diverges, Kummer's divergent half gives divergence of ∑an\sum a_n (K1=−0.75K_1 = -0.75, K100=−51101≈−0.505K_{100} = \frac{-51}{101} \approx -0.505 — matching the caption).

6. Series